What Velocity Does the Boy Gain After Throwing the Jug?

[GodAllen]

Homework Statement

A boy on roller blades throws a jug of water away from himself, giving it a speed of 14m/s. The boy's mass is 38kg and the mass of he jug and the water is 7.9kg. What is the velocity gained by the lad?

Homework Equations

Not sure.

The Attempt at a Solution

Not sure.

 

[dazedy]

This is a simple conservation of momentum problem, so assume that at the beginning, total momentum is equal to 0
Afterwards, the total momentum must stay the same.
Thus:
Momentum(before) = Momentum(boy) + Momentum(jug)
This should be enough to get you started

 

[mido808]

so you apply the formula P(before)=P(after). (And by the way the momentum before is 0 as the intial velocities of the boy and the jug are 0 as P(before)= m(boy)xv(boyintial)+m(jug)xv(jug int.))Hope u understood until now.
p(after)=m(boy)xv(boy final)+m(jug)xv(jug)
P(after)=38kg(v)+7.9kg(14m/s)
P(after)=38v+110.6
P(before)=p(after)
0=38v+110.6
38v=-110.6
V=-2.9
So velocity of the boy is 2.9 and its direction is in the opposite direction that he threw the jug in.
Thanks for reading in my answer and i hope u understand and everbody as well since i am a new user.

 

[ashishsinghal]

so you apply the formula P(before)=P(after). (And by the way the momentum before is 0 as the intial velocities of the boy and the jug are 0 as P(before)= m(boy)xv(boyintial)+m(jug)xv(jug int.))Hope u understood until now.
p(after)=m(boy)xv(boy final)+m(jug)xv(jug)
P(after)=38kg(v)+7.9kg(14m/s)
P(after)=38v+110.6
P(before)=p(after)
0=38v+110.6
38v=-110.6
V=-2.9
So velocity of the boy is 2.9 and its direction is in the opposite direction that he threw the jug in.
Thanks for reading in my answer and i hope u understand and everbody as well since i am a new user.

you are not supposed to solve the whole problem. this is against pf rules
https://www.physicsforums.com/showthread.php?t=414380